Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 1 / 6C / Solution

If $A{x}_p=b$, then $Ax=b$ is equivalent to $A(x-x_p)=0$. Thus the solution set is the affine solution space of a linear equation $\boxed{x_p+\ker A}$, and every vector of that form solves the equation.

For the symmetric matrix case, let $A{u}_i={\lambda }_i{u}_i$. Symmetry givesDistinct eigenvalues therefore imply $u_i\cdot u_j=0$. Normalizing the three vectors gives an orthonormal basis. Expand $x={\sum }_i{t}_i{u}_i$ and $b={\sum }_i{b}_i{u}_i$, with $b_i=b\cdot u_i$. The equation $(A-{\lambda }_{k}I)x=b$ becomes $({\lambda }_i-{\lambda }_k)t_i=b_i$ for every $i$. Thus solvability is equivalent to $b_k=0$, and all solutions areThis proves the claimed orthogonality condition and the full solution family rather than only a particular solution.

For the numerical symmetric matrix, direct multiplication gives $A{u}_1=0$ for $u_1=(1,1,1{)}^T/\sqrt{3}$, and $A{u}_2=-\sqrt{2}{u}_2$ for $u_2=(1,-1,0{)}^T/\sqrt{2}$. Their orthogonal complement is spanned by $u_3=(1,1,-2{)}^T/\sqrt{6}$, and multiplication gives $A{u}_3=\sqrt{3}{u}_3$. The supplied right side has $b\cdot u_1=0$, so the linear system is solvable. Its other coefficients are $b\cdot u_2=2$ and $b\cdot u_3=3\sqrt{2}$. HenceEquivalently, all solutions are